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Unit 3: Systems of Linear Equations
Lesson 1: Solving Systems by Graphing
Lesson 2: Solving Systems by Substitution
Lesson 3: Solving Systems by Elimination
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3.1
Solving Systems by Graphing
Lesson Objectives:
situation
graphs
Create a system of linear equations to model a
Solve systems of linear equations using tables and
Consider This…
In Unit 2 you gained experience in writing linear
equations with two variables. Sometimes,
problems involve two linear equations that have
to be solved simultaneously (the solution for one
equation must also be a solution for the other
equation). The task is to find one pair (x, y) of
values that satisfies both linear equations.
Students in the Hamilton High School science
club faced this kind of problem when they tried
to raise $240 to buy a special eyepiece for their
high-powered telescope at their school. The
school PTA offered to pay club members for an
after school work project that would clean up a
nearby park and recreation center building.
Because the outdoor work was harder and dirtier, the deal with the PTA would pay $16 for each
outdoor worker and $10 for each indoor worker. The club had 18 members eager to work on the
project.
The conditions in the science club’s situation could be represented by linear equations like these:
16x + 10y = 240 and x + y = 18
1. What do the x and the y represent in terms of this situation?
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2. What problem condition does the equation 16x + 10y = 240 represent?
3. What problem condition is represented by x + y = 18?
4. What are two combinations of outdoor and indoor workers that will allow the club to earn
just enough money to buy the telescope eyepiece?
5. Will either of these combinations also put each willing club member to work? If not, do you
think there is such a combination and why?
6. Graph each of the equations in this situation in a different color. Label each axis and identify
the scale that you are using.
7) Does the graph help you identify a “simultaneous” solution, and if so, what do you estimate
that solution to be?
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Summarize
 A "system" of equations is a set or collection of equations that you deal with all together at
once. Linear equations (ones that graph as straight lines) are simpler than non-linear equations, and
the simplest linear system is one with two equations and two variables.
Now consider the following two-variable system of linear equations:
y = 3x – 2
y = –x – 6
Since the two equations above are in a system, we deal with them together at the same time. In
particular, we can graph them together on the same axis system, like this:
A solution for a single equation is any point that lies on the line for that equation. A solution for a
system of equations is any point that lies on each line in the system.
For example, the point shown below is
not a solution to the system,
because it is not on either line:
The point shown in the graph below is also not a solution
to the system, because it lies on only one of the lines,
not on both of them:
The point shown below IS a solution to the system,
because it lies on BOTH of the lines:
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This point marks the intersection of
the two lines. Since this point is on
both lines, it thus solves both
equations, so it solves the entire
system of equation. And this
relationship is always true:
For systems of equations,
"solutions" are "intersections".
 You can confirm the solution by plugging it into the system of equations, and confirming that
the solution works in each equation.
Checking your solution:
Determine whether either of the points (–1, –5) and (0, –2) is a solution to the given system of
equations.
y = 3x – 2
y = –x – 6
To check the given possible solutions, plug the x- and y-coordinates into the equations, and check to
see if they work.
Checking (–1, –5):
(–5) ?=? 3(–1) – 2
–5 ?=? –3 – 2
–5 = –5 (solution checks)
(–5) ?=? –(–1) – 6
–5 ?=? 1 – 6
–5 = –5 (solution checks)
Since the given point works in both equations, it IS a solution to the system.
Now I'll check the other point (which we already know, from looking at the graph, is not a solution):
Checking (0, –2):
(–2) ?=? 3(0) – 2
–2 ?=? 0 – 2
–2 = –2 (solution checks)
So the solution works in one of the equations. But to solve the system, it has to work
in both equations. Continuing the check:
(–2) ?=? –(0) – 6
–2 ?=? 0 – 6
–2 ?=? –6
But –2 does not equal –6, so this "solution" does not check. Then the answer is:
Only the point (–1, –5) is a solution to the system.
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Solving a System using Graphs
If you can graph a straight line, you can solve systems of equations graphically!
The process is very easy. Simply graph the two lines and look for the point where they intersect
(cross).
Solve graphically:
4x - 6y = 12
2x + 2y = 6
To solve a system of equations graphically, graph both equations and see where they intersect. The
intersection point is the solution.
First, solve EACH equation for "y =" :
4x - 6y = 12
4x=6y + 12
4x-12=6y
y=4x/6 – 12/6
y=2/3x-2
2x + 2y = 6
2x+2y=6
2y=-2x+6
y=-2x/2 + 6/2
y=-1x+3
slope =2/3
y-intercept = -2
slope = -1
y-intercept = 3
Graph the lines:
The point of intersection of the two lines, (3,0), is the
solution to the system of equations.
This means that (3,0), when substituted into either
equation, will make them both true.
Check: Since the two lines cross at (3,0), the solution is
x =3 and y = 0. Checking these value shows that this answer
is correct. Plug these values into the ORIGINAL equations
and get a true result.
4x - 6y = 12 2x + 2y = 6
4(3) - 6(0) = 12
12 - 0 = 12
12 = 12
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2(3) + 2(0) = 6
6+0=6
6=6
Practice
DIRECTIONS: Solve the following systems of equations by making a table and graphing the
equation. If the equation is not in slope-intercept (y=mx+b) form you should solve for “y” first!
Write your solution as an ordered pair. You do not have to represent every pair in the table
on your graph.
1) y1 = 6 – 3x
y2 = 2x + 1
y1
y2
-2
-1
0
1
2
3
4
5
Solution ________________________
Check:
2) y1 = 4x – 2
y2 = x + 1
x
-2
-1
0
1
2
3
4
5
y1
y2
Solution ________________________
Check:
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3) 3y = 18x - 6
x + y =5
x
-2
-1
0
1
2
3
4
5
y1
y2
Solution ________________________
Check:
4) x - 5y = -11
4x + 2y = 22
x
-2
-1
0
1
2
3
4
5
y1
y2
Solution ________________________
Check:
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Summarize
When you are solving systems, you are, graphically, finding intersections of lines. For two-variable
systems, there are then three possible types of solutions:
The first graph (to the right) shows two distinct non-parallel lines that
cross at exactly one point. This is called an "independent" system of
equations, and the solution is always some (x,y) point.
The second graph (to the left) shows two distinct lines that are parallel.
Since parallel lines never cross, then there can be no intersection; that is, for
a system of equations that graphs as parallel lines, there can be no solution.
This is called an "inconsistent" system of equations, and it has no
solution.
The third graph (to the right) appears to show only one line. Actually, it's
the same line drawn twice. These "two" lines, really being the same line,
"intersect" at every point along their length. This is called a "dependent"
system, and the "solution" is the whole
Practice
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DIRECTIONS: Solve the following systems of equations by graphing. If the equation is not in
slope-intercept (y=mx+b) form you should solve for “y” first! Write your solution as an ordered
pair.
1) 4x – y = -3
2y – 4x = 2
Solution ________________________
2) 2x + y = 5
2x + y = 1
Solution ________________________
3) - 2x + y = 3
- 4x + 2y = 6
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Solution ________________________
Apply
4) Suppose that one music subscription service, Lisa Jams, charges $11 per month plus $0.85
per song; and a second service, Amy Rocks, charges $8 per month plus $1.00 per song. Let C
be the monthly cost in dollars and n be the number of songs downloaded in a month.
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a.
Write a linear equation that represents the monthly subscription cost for Lisa Jam’s.
b. Write a linear equation that represents the monthly subscription cost for Amy Rock’s.
c. Compare the monthly subscription cost for each company if you download 10 songs.
Which is cheaper?
d. Compare the monthly subscription cost for each company if you download 30 songs.
e. How many songs could you download from each if you had $50?
f. Create a system of equations that will allow you to find the number of songs and cost at
which the two services are the same.
(continued)
Write your system of equations here:
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g. Graph each of the equations, clearly labeling the x- and y- axis. Graph each equation in a
different color.
h. For what number of songs do Lisa Jams and Amy Rocks have the same cost? What is that cost?
Check your solution.
Apply
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5) The total activity fees at Suffolk College is calculated by charging $25 per activity plus an
additional flat fee of $30 for the semester. Boston College charges $35 per activity with no
additional fee.
a. Write a system of equations that represents the relationship between the amount of
activities, a, and the total cost of fees, c.
b. Graph each of the equations, clearly labeling the x- and y-axis. Graph each
equation in a different color.
c. What is the solution to this system of equations? Check your solution.
d. What is the real world meaning of the solution?
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3.2
Solving Systems by Substitution
Lesson Objectives:
Solve a system of
linear equations using substitution
Develop equations for given situations and locate
the point of intersection for the situation by substitution
Consider This…
Nicole has $100 in the bank, but ends up spending it at a rate of $3 each day. Billy has $20 in the
bank, but is adding to it by saving $5 each day.
1. Write an equation that shows the amount of money left in the bank account, y, as a
function of the day, x, for Billy’s bank account.
2. Write an equation that shows the amount of money left in the bank account, y, as a
function of the day, x, for Nicole’s bank account.
3. How can we combine the two equations listed above into one equation with one variable?
(What do you notice about the y in both cases?)
4. Solve for the number of days where both Nicole and Billy will have the same amount of
money. How much money do they have in their account?
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Summarize
To solve a system by SUBSTITUTION:

When we solve a system, we are finding the coordinates, (x, y), that satisfy both equations. In
order to find these points by SUBSTITUTION, there are a few steps to follow:
Example:
10y = 5000 + 100x
2y = 2000 + 10x
1. Isolate one variable from either equation
10y = 5000 + 100x becomes
y= 500+ 10x
2. Substitute the expression that is equivalent to
that variable into the other equation
2(500+10x)= 2000 + 10x
3. Solve the equation
1000 + 20x= 2000 + 10x
10x = 1000
x= 100
4. Substitute the value into one of the original
equations and solve.
y = 500 + 10 (100)
y= 500+ 1000
y=1500
5. Now you have both coordinates that satisfy
both equations.
x = 100 and y = 1500
(100, 1500)
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Practice
DIRECTIONS: Solving each system of equations using the substitution method. Show all steps.
1. y = -4x + 32
y=x–3
2. d = r + 1
d = -2r – 2
3. a + b = 11
3a – 2b = 8
4. 4d – 3h = 25
3d – 12h = 9
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5. 2x = 3y
4x – 3y = 12
7.
7x – 3y = 23
x + 2y = 13
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6.
2x + 3y = 7
4x – 5y = 25
8.
x = 4y
2x + 3y = 22
Summarize
 When solving a system by substitution, we can still have two lines that never intersect, which
results in NO SOLUTION or two lines that are exactly the same equations, which results in INFINITELY
MANY SOLUTIONS
NO SOLUTION
The equations have the same
slope, but a different y-intercept
INFINITELY MANY
The equations have the same slope,
AND the same y-intercept
EXAMPLE:
EXAMPLE:
Since the slope of these equations
are both =- 1 and they have
different y-intercepts, we know this
will have no solution. If we solve,
this is what happens:
Since the slope of these equations
both = 3/2 and they have the same
y-intercept of -2, we know this will
have no solution. If we solve, this is
what happens:
x+y=3y=3–x
3x – 2y = 4 y = (3/2)x -2
x + (3-x) = -2
x+ 3 – x = -2
3= - 2
6x – 4((3/2)x – 2) = 8
6x – 6x + 8 = 8
8 =8
The variables cancel and we are left
with an UNTRUE statement.
The variables cancel and we are left
with a TRUE statement.
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Practice
DIRECTIONS: Solve each of these systems using the substitution method. State the point of
intersection or write “no solution” or ”infinitely many solutions.”
1.
3.
-2x – y = 3
x + 2y = 4
2.
-2x + y = 3
-4x +2y = 2
y = -4x + 5
y = -4x – 2
4.
y = 1.5x + 9
y = (3/2)x + 9
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5. x + 2y = 8
2x + y = 4
7.
6.
3x – 2y = 1
6x – 4y = 10
8.
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x – 3y = 6
3x – 9y = 18
x + 7 = 10
2x – 5y = 16
Summarize
In order to solve a word problem by substitution, you must come up with both equations first, then
follow the steps on the other page.
EXAMPLE: For the school play, adult tickets, a, sell for $5 and student tickets, s, sell for $3. On
Saturday, a total of 390 tickets were sold and the total receipts were $1530. Unfortunately, no one
knows how many of the tickets were adult and how many were student. Fortunately, you’re on the
job and can figure it out.
To set up the system:
1. Look at the two variables identified
2. You should be given two different pieces
on information about the variables.
These are the equations that you will be
using.
3. Set up the equations and solve these
equations by following the method on
the previous page.
We have adult and student tickets, a and s.
We know the TOTAL COST and the TOTAL
PEOPLE attending.
TOTAL COST: 1530 = 5a + 3s
( $5 per adult, $3 per student)
TOTAL PEOPLE: 390 = a + s
To Solve:
1530 = 5a + 3s

 390 = a + s
rearrange
substitute
solve
 a = 390 – s  1530 = 5(390 – s) + 3s  1530 = 1950 -5s +3s
1530 – 1950 = -2s
-420 = -2s
210 = s
To find a:
a = 390 – (210)
a = 180
Therefore, in order to sell 390 tickets and earn $1530, 180 adult tickets and 210 student tickets must
be sold.
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Practice
DIRECTIONS: ome up the systems for the following word problems. Use substitution to solve. Be
sure to write a statement that answers the question directly.
1. Michael buys two bags of chips and three boxes of pretzels for $5.13. He then buys another bag of
chips and two more boxes of pretzels for $3.09. Find the cost of each bag of chips and each box of
pretzels.
2. A test has twenty questions worth 100 points. The test consists of True/False questions worth 3
points each and multiple choice questions worth 11 points each. How many multiple-choice
questions are on the test?
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3. If 4 apples and 2 oranges equals $1 and 2 apples and 3 orange equals $0.70, how much does each
apple and each orange cost?
4. A soccer team bought ice cream cones to celebrate a victory. The total cost of 12 double cones
and 8 single cones was $17. A double cone cost $0.25 more than a single cone. What is the price of
each type of cone?
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Apply
5. The senior class of Shrewsbury is planning the hypnotist show to raise money for their class. Based
on previous fundraising events, they would like to get 300 people in the auditorium. One plan is to
charge adults $8 and students $3 admission. The club wants to earn $2,000 from admission charges.
a. What is the equation that would represent the number of people, a and s?
b. What is the equation that would represent the profit?
c. How many adults and students need to attend in order to earn the $2000 and have 300
people?
d. If the cost was $10 for adults and $5 for students, how would that change the number of
adults and students attending in order to reach their goal? Show or explain your work.
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3.3
Solving Systems by Elimination
Lesson Objectives:
Understand that
multiplying both sides of a linear equation does not change the
values of x and y that satisfy the equation
Use this idea to solve systems by combining the
two equations to eliminate one of the variables
Consider This…
In Lesson 2, we looked at the Hypnotist Show
put on by the Senior Class at Shrewsbury High
School to raise money.
We came up with the system of equations
a+ s =300
8a+3s=2000
where a represented the number of adults
tickets sold and s represented the number of
student tickets sold. Solving this system told us
that the Senior Class should try to sell 220 adult
tickets and 80 student tickets in order to reach
their goal of making $2000.
1. The Senior Class decides that they should actually try to double their original income goal of
$2000, but the auditorium where the show is being held has a capacity of 300 seats, so they
can’t sell any more tickets. One student suggests doubling the price per ticket for adults and
students. Write a system of equations that models this new situation.
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2. Does the original solution of selling 220 adult tickets and 80 student tickets satisfy this system
of equations? Show your work.
3. Another student suggests a different way to reach an income goal of $4000. She thinks the
class should move the show to the Commons, which has a capacity of 600 people, but that the
class should keep the ticket prices the same (that is, $8 per adult ticket and $3 per student
ticket). Write a system of equations that models this new situation.
4. Does the original solution of selling 220 adult tickets and 80 student tickets satisfy this system
of equations? Show your work.
5. Using either graphing or substitution, solve the system you came up with in #3 (above) to find
out how many adult tickets and how many student tickets the class would need to sell in
order to meet their income goal of $4000 with 600 total tickets. Show your work.
6. What do you notice about your solution to #5 and the
original solution?
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Summarize
 Multiplying (or dividing) both sides of an equation does NOT change the values that satisfy the equation.
For example, when we used the first student’s suggestion of doubling the price per ticket for
adults and students in order to double the profit, we created the system:
a + s = 300
2(8a+3s = 2000)
 a + s = 300
 16a+3s= 4000 This still had the same solution of (220, 80).
However, multiplying (or dividing) only one side an equation DOES change the values that satisfy the equation.
For example, when we used the second student’s suggestion of just doubling the number of
tickets to sell in order to double the profit, we created the system:
a + s = (300)2
8a+3s = (2000)2
 a + s = 600
 8a+3s= 4000 This had a different solution of (440, 160).
We can use this idea to help us solve systems of linear equations, and we call
this method of solving combination or elimination.
I want to
eliminate s
TO SOLVE A SYSTEM USING COMBINATION/
ELIMINATION:
1st Look at your two equations and select one variable
that you want to eliminate.
a + s = 300
8a +3s = 2000
I see that one s has a +3
coefficient, so I know that
the other s needs a -3
2nd Figure out what coefficients would make that variable
“cancel out”
3rd Multiply both sides of one or both equations the make the
coefficients of your chosen variable equal opposites.
-3( a + s = 300)  -3a + -3s =-900
8a +3s = 2000

8a + 3s = 2000
4th Combine the two equations by adding them
together to get one equation with only one variable
I chose to multiply the top
equation by -3. I didn’t need
to multiply the bottom
equation.
-3a + -3s =-900
8a + 3s = 2000
5a + 0 = 1100
5a + 0 = 1100 variable
5a = 1100
a = 220
5th Solve this new equation for the remaining
a + s = 300
220 + s = 300
s = 80
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6th Substitute this value into either original equation, and solve for the other variable (that
you initially eliminated)
Summarize
Example:
Solve the system
4d - 3h = 25
3d -12h = 9 by combination/ elimination.
1st I need to decide what variable to eliminate- I will choose to eliminate h.
2nd In order to eliminate h, I need the coefficients of the h terms to be equal opposites.
Since the h in the first equation has a -3 coefficient and the h in the second equation
has a -12 coefficient, and I know that 3 goes into 12, I will try to make the coefficient of
h in the first equation into +12.
3rd I will do this by multiplying both sides of the first equation by -4 (because -3x-4=12)
-4(4d - 3h = 25)  -16d + 12h = -100
3d -12h = 9
 3d - 12h = 9
4th I want the +12h and the -12h to cancel out. This will happen when I add the two
equations together.
-16d + 12h = -100
+
3d - 12h = 9
-13d + 0 = -91
5th Now I can solve this new equation for d.
-13d + 0 = -91
-13d = -91
d=7
6th Since I know that d=7, I can use this to find h by substituting into one of the original
equations and solving for h.
4d - 3h = 25
4(7) - 3h = 25
28 - 3h = 25
- 3h = -3
h=1
So, my solution is (d, h) = (7, 1).
I should check my work, too, by substituting d=7 and h=1 into both equations:
32
4d - 3h = 25
4(7) - 3(1) = 25
28 - 3 = 25
25 = 25
3d - 12h = 9
3(7) -12(1) = 9
21 - 12 = 9
9=9
Summarize
Example:
Solve the system
3x - 2y = 21
4x + 3y =11
by combination/ elimination.
1st I need to decide what variable to eliminate- I will choose to eliminate y.
2nd In order to eliminate y, I need the coefficients of the y terms to be equal opposites.
Since the y in the first equation has a -2 coefficient and the y in the second equation
has a +3 coefficient, I know that I need to find a number that both 2 and 3 are factors
of (that is, multiply into). Since I know that 2×3 = 6, I will try to make the coefficient of
y in both equations into 6.
3rd I will do this by multiplying both sides of the first equation by 3 and both sides of the
second equation by 2.
3(3x - 2y = 21)  9x - 6y = 63
2(4x + 3y = 11)  8x + 6y = 22
4th I want the -6y and the +6y to cancel out.
equations together.
This will happen when I add the two
9x - 6y = 63
+ 8x + 6y = 22
17x + 0 = 85
th
5 Now I can solve this new equation for x.
17x + 0 = 85
17x = 85
x = 5
6th Since I know that x=5, I 4x + 3y = 11 can use this to find y by substituting into
one of the original equations 4(5) + 3y = 11 and solving for y.
20 + 3y = 11
3y = -9
y = -3
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So, my solution is (x, y) = (5, -3)
I should check my work, too, by substituting x=5 and y=-3 into both equations:
4x + 3y = 11
4(5) + 3(-3) = 11
20 + -9 = 11
11 = 11
3x - 2y = 21
3(5) -2(-3) = 21
15 + 6 = 21
21 = 11
Practice
DIRECTIONS: Solve each of the following systems using combination/ elimination. SHOW ALL
STEPS, and check your final answer. Remember, a system of linear equations can have one
solution, no solutions, or infinitely many solutions.
1.
3.
x+y=3
x - y = -1
3x - y = -4
3x - y = 0
2.
2x - y = -3
x + y = -3
4.
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2x - y = 3
4x - 2y = 6
35
5.
a - 4b = 8
2a - 8b = 16
7.
p+r=2
p - r = -4
6.
8.
36
3h - d = 4
2h - 3d = -9
2m - 6n = -24
m - 3n = 18
Apply
9. To raise money for a trip to Europe over April vacation, a travelling basketball team put on a
“3-on-3” Basketball Tournament, with special appearances from local celebrities. The entry
fee for a player was $15, and the entry fee for a spectator was $5. However, security only
kept track of the total number of people who entered the event, and reported that this
number was 152. The organizers counted that they had $1300 from admissions.
a. Write a system of equations that will help you figure out how many players, x, and
spectators, y, attended the tournament.
b. Solve this system using the combination/elimination method. SHOW ALL STEPS.
c. Check your solution in your original system.
d. Write a sentence about what your solution means in the context of the problem.
37
Apply
10. Dom, Tom, and their mom go to Moe’s® Southwest Grill for lunch. Dom orders 3 hard tacos
and a quesadilla, and his order costs $9.16. Tom orders 2 hard tacos and 2 quesadillas, and his
order costs $10.76. (They both, of course, enjoy the free tortilla chips and salsa that come with
every order.)
a. Write a system of equations that will help you figure out the price
per hard taco, t, and the price per quesadilla, q.
b. Solve this system using the combination/ elimination method.
c. Check your solution in your original system.
d. Some of Dom and Tom’s football teammates join them for lunch. Between them, they
order 22 hard tacos and 16 quesadillas. How much does their order cost? Show your
work.
38
Apply
11. Your “Focus on Foods” class is baking pumpkin bread and banana bread for this week’s
lesson. Your teacher has only 36 cups of flour and 12 ½ cups of sugar in her pantry. Each
loaf of banana bread requires 3 cups of flour and ½ cup of sugar, while each loaf of pumpkin
bread requires 2 ½ cups of flour and 1 ½ cups of sugar.
a. Write a system of equations that will help you figure out the number of loaves of
pumpkin bread, p, and the number of loaves of banana bread, b, that your class can
bake this week.
b. Solve this system using any method that we have learned. SHOW YOUR WORK.
(If you solve by graphing, you must draw your graphs below.)
c. Check your solution in your original system.
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Unit 3
Reflect & Review
You should be able to…
Create a system of linear
equations to model a situation
Understand what is means to solve a system, in terms of the
equations and in terms of the graph
Solve a system of linear equations using tables
Solve a system of linear equations using graphs
Solve a system of linear equations using substitution
Understand that multiplying both sides of a linear equation
does not change the values of x and y that satisfy the
equations
Use the above idea to solve systems by combining the two
equations to eliminate one of the variables (i.e. solve a system
of linear equations using elimination)
Find the point of intersection of two lines by solving a system
of linear equations using substitution or elimination
Solve real-world problems using systems of linear equations
(write and solve a system given a verbal description)
Vocabulary
System of equations
Solution to a system of equations
Intersection
No solution
Infinitely many solutions
Substitution method
Combination/ elimination method
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